Rate of Escape of the Mixer Chain

0.17 MB·English

by  Ariel Yadin

9 Rate of Escape of the Mixer Chain
0
0
2
∗
Ariel Yadin
n
a
J

3
1
Abstract

] The mixer chain on a graph G is the following Markov chain. Place tiles on the
R
P verticesof G,each tilelabeled byitscorresponding vertex. A“mixer” movesrandomly
. on the graph, at each step either moving to a randomly chosen neighbor, or swapping
h
t thetile at its current position with some randomly chosen adjacent tile.
a
m We study the mixer chain on Z, and show that at time t the expected distance to
[ theorigin is t3/4,up to constants. This is a new exampleof arandom walk on a group

with rate of escape strictly between t1/2 and t.
2
v
9
2
1 Introduction
1
6
0
5 Let G = (V,E) be a graph. On each vertex v V, place a tile marked v. Consider the
∈
0 following Markov chain, which we call the mixer chain. A “mixer” performs a random walk
/
h on the graph. At each time step, the mixer chooses a random vertex adjacent to its current
t
a position. Then, with probability 1/2 it moves to that vertex, and with probability 1/2 it
m
remains at the current location, but swaps the tiles on the current vertex and the adjacent
:
v
vertex. If G is the Cayley graph of a group, then the mixer chain turns out to be a random
i
X
walk on a different group.
r
a Aside from being a canonical process, the mixer chain is interesting because of its rate of
escape. Fora randomwalk X onsomegraphG, we use the terminologyrate of escape for
t
{ }
the limit
logE[d(X ,X )]
t 0
lim ,
t→∞ logt
whered(, )isthe graphicaldistance. Whenrestrictingto randomwalksongroups,itisstill
· ·
open what values in [0,1] can be obtained by rates of escape. For example, if the group is
∗Faculty of Mathematics and Computer Science, The Weizmann Institute of Science, Rehovot 76100,
Israel. Email: [email protected]
1
Zd then the rate of escape is 1/2. On a d-ary tree (free group) the rate of escape is 1. As
far as the author is aware,the only other examples known were given by Erschler in [1] (see
also [2]). Erschler iterates a construction known as the lamp-lighter (slightly similar to the
mixer chain), and produces examples of groups with rates of escape 1 2−k, k =1,2,...,.
−
After formally defining the mixer chain on general groups, we study the mixer chain on Z.
Our main result, Theorem 2.1, shows that the mixer chain on Z has rate of escape 3/4.
It is not difficult to show (perhaps using ideas from this note) that on transient groups the
mixer chain has rate of escape 1. Since all recurrent groups are essentially Z and Z2, it
seems that the mixer chain on other groups cannot give examples of other rates of escape.
As for Z2, one can show that the mixer chain has rate of escape 1. In fact, the ideas in this
note suggest that the distance to the origin in the mixer chain on Z2 is nlog−1/2(n) up to
constants.
Afterintroducingsomenotation,weprovideaformaldefinitionofthemixerchain,asrandom
walk on a Cayley graph. The generalizationto general graphs is immediate.
Acknowledgement. I wish to thank Itai Benjamini for suggesting this construction, and
for useful discussions.
1.1 Notation
Let G be a group and U a generating set for G, such that if x U then x−1 U (U is
∈ ∈
called symmetric). The Cayley graph of G with respect to U is the graph with vertex set
G and edge set g,h : g−1h U . Let be a distribution on U. Then we can define
{ } ∈ D
the random walk on G (with respect to U and ) as the Markov chain with state space G
(cid:8) (cid:9)
D
and transition matrix P(g,h)= 1 g−1h U (g−1h). We follow the convention that such
∈ D
a process starts from the identity element in G.
(cid:8) (cid:9)
A permutation of G is a bijection from G to G. The support of a permutation σ, denoted
supp(σ), is the set of all elements g G such that σ(g) = g. Let Σ be the group of
∈ 6
all permutations of G with finite support (multiplication is composition of functions). By
< g,h > we denote the transposition of g and h; that is, the permutation σ with support
g,h such that σ(g)=h, σ(h)=g. By <g ,g ,...,g > we denote the cyclic permutation
1 2 n
{ }
σ with support g ,...,g , such that σ(g )=g for j <n and σ(g )=g .
1 n j j+1 n 1
{ }
For an element g G we associate a canonical permutation, denoted by φ , defined by
g
∈
φ (h)=gh for all h G. It is straightforwardto verify that the map g φ is a homomor-
g g
∈ 7→
2
phismofgroups,andso we use g to denote φ . Althoughg Σ, we havethatgσg−1 Σ for
g
6∈ ∈
all σ Σ.
∈
We now define a new group, that is in fact the semi-direct product of G and Σ, with respect
to the homomorphism g φ mentioned above. The group is denoted by G⋉Σ, and its
g
7→
elements are G Σ. Group multiplication is defined by:
×
(g,σ)(h,τ)d=ef (gh,gτg−1σ).
We leave it to the reader to verify that this is a well-defined group operation. Note that the
identity element in this group is (e,id), where id is the identity permutation in Σ and e is
the identity element in G. Also, the inverse of (g,σ) is (g−1,g−1σ−1g).
We use d(g,h) = d (g,h) to denote the distance between g and h in the group G with
G,U
respect to the generating set U; i.e., the minimal k such that g−1h = k u for some
j=1 j
u ,...,u U. The generating set also provides us with a graph structure. g and h are
1 k Q
∈
said to be adjacent if d(g,h) = 1, that is if g−1h U. A path γ in G (with respect to the
∈
generating set U) is a sequence (γ ,γ ,...,γ ). γ denotes the length of the path, which is
0 1 n
| |
defined as the length of the sequence minus 1 (in this case γ =n).
| |
1.2 Mixer Chain
In order to define the mixer chain we require the following
Proposition 1.1. Let U be a finite symmetric generating set for G. Then,
Υ= (u,id),(e,<e,u>) : u U
{ ∈ }
generates G⋉Σ. Furthermore, for any cyclic permutation σ =<g ,...,g > Σ,
1 n
∈
n
dG⋉Σ,Υ((g1,σ),(g1,id)) 5 d(gj,σ(gj)).
≤
j=1
X
Proof. Let D((g,σ),(h,τ)) denote the minimal k such that (g,σ)−1(h,τ) = k υ , for
j=1 j
some υ ,...,υ Υ, with the convention that D((g,σ),(h,τ)) = if there is no such
1 k Q
∈ ∞
finite sequence of elements of Υ. Thus, we want to prove that D((g,σ),(e,id)) < for all
∞
g G and σ Σ. Note that by definition for any f G and π Σ, D((g,σ),(h,τ)) =
∈ ∈ ∈ ∈
D((f,π)(g,σ),(f,π)(h,τ)).
3
A generator simple path in G is a finite sequence of generators u ,...,u U such that for
1 k
∈
any 1 ℓ k, k u = e. By induction on k, one can show that for any k 1, and for
≤ ≤ j=ℓ j 6 ≥
any generator simple path u ,...,u ,
Q 1 k
k k−1 k−1
(e,<e, u >)= (e,<e,u >)(u ,id) (e,<e,u >) (e,<e,u−1 >)(u−1 ,id).
j j j · k · k−j k−j
j=1 j=1 j=1
Y Y Y
(1.1)
If d(g,h) = k then there exists a generator simple path u ,...,u such that h = g k u .
1 k j=1 j
Thus, we get that for any h G,
∈ Q
D((e,<e,h>),(e,id)) 4d(h,e) 3.
≤ −
Because g <e,g−1h>g−1 =<g,h>, we get that if τ =<g,h>σ then
D((g,τ),(g,σ))=D((g,σ)(e,<e,g−1h>),(g,σ)(e,id)) 4d(g−1h,e) 3=4d(g,h) 3.
≤ − −
The triangle inequality now implies that D((h,τ),(g,σ)) 5d(g,h) 3.
≤ −
Thus, if σ =<g ,g ,...,g >, since σ =<g ,g ><g ,g > <g ,g >, we get that
1 2 n 1 2 2 3 n−1 n
···
n−1
D((g ,σ),(g ,id)) 5 d(g ,g )+d(g ,g ). (1.2)
1 1 j j+1 n 1
≤
j=1
X
The propositionnow follows from the fact that any σ Σ can be written as a finite product
∈
of cyclic permutations.
⊓⊔
We are now ready to define the mixer chain:
Definition 1.2. Let G be a groupwith finite symmetric generatingset U. The mixer chain
on G (with respect to U) is the random walk on the group G⋉Σ with respect to uniform
measure on the generating set Υ= (u,id),(e,<e,u>) : u U .
{ ∈ }
An equivalent way of viewing this chain is viewing the state (g,σ) G ⋉ Σ as follows:
∈
The first coordinate corresponds to the position of the mixer on G. The second coordinate
corresponds to the placing of the different tiles, so the tile marked x is placed on the vertex
σ(x). By Definition 1.2, the mixer chooses uniformly an adjacent vertex of G, say h. Then,
withprobability 1/2the mixer swapsthe tiles onh andg, andwith probability1/2it moves
to h. The identity element in G⋉Σ is (e,id), so the mixer starts at e with all tiles on their
corresponding vertices (the identity permutation).
4
1.3 Distance Bounds
In this section we show that the distance of an element in G⋉Σ to (e,id) is essentially
governedby the sum of the distances of each individual tile to its origin.
Let (g,σ) G⋉Σ. Let γ = (γ ,γ ,...,γ ) be a finite path in G. We say that the path γ
0 1 n
∈
covers σ if supp(σ) γ ,γ ,...,γ . The covering number of g and σ, denoted Cov(g,σ),
0 1 n
⊂{ }
is the minimal length of a path γ, starting at g, that covers σ; i.e.
Cov(g,σ)=min γ : γ =g and γ is a path covering σ .
0
{| | }
To simplify notation, we denote D =dG⋉Σ,Υ.
Proposition 1.3. Let (g,σ) G⋉Σ. Then,
∈
D((g,σ),(g,id)) 2Cov(g,σ)+5 d(h,σ(h)).
≤
h∈sXupp(σ)
Proof. The proof of the proposition is by induction on the size of supp(σ). If supp(σ) =0,
| |
then σ =id so the proposition holds. Assume that supp(σ) >0.
| |
Let n=Cov(g,σ), and let γ be a path in G such that γ =n, γ =g and γ covers σ. Write
0
| |
σ = c c c , where the c ’s are cyclic permutations with pairwise disjoint non-empty
1 2 k j
···
supports, and
k
supp(σ)= supp(c ).
j
j=1
[
Let
j =min m 0 : γ supp(σ) .
m
{ ≥ ∈ }
So, there is a unique 1 ℓ k such that γ supp(c ). Let τ =c−1σ. Thus,
≤ ≤ j ∈ ℓ ℓ
supp(τ)= supp(c ),
j
j6=ℓ
[
and specifically, supp(τ) < supp(σ). Note that h supp(γ−1c γ ) if and only if γ h
| | | | ∈ j ℓ j j ∈
supp(c ), and specifically, e supp(γ−1c γ ). γ−1c γ is a cyclic permutation, so by Propo-
ℓ ∈ j ℓ j j ℓ j
sition 1.1, we know that
D((γ ,σ),(γ ,τ))=D((γ ,τ)(e,γ−1c γ ),(γ ,τ))=D((e,γ−1c γ ),(e,id))
j j j j ℓ j j j ℓ j
5 d(γ−1h,γ−1c (h))=5 d(h,σ(h)). (1.3)
≤ j j ℓ
h∈sXupp(cℓ) h∈sXupp(cℓ)
5
By induction,
D((γ ,τ),(γ ,id)) 2Cov(γ ,τ)+5 d(h,τ(h)). (1.4)
j j j
≤
h∈sXupp(τ)
Let β be the path (γ ,γ ,...,γ ). Since γ is the first element in γ that is in supp(σ), we
j j+1 n j
get that supp(τ) supp(σ) γ ,γ ,...,γ , which implies that β is a path of length
j j+1 n
⊂ ⊆ { }
n j that covers τ, so Cov(γ ,τ) n j. Combining (1.3) and (1.4) we get,
j
− ≤ −
D((g,σ),(g,id)) D((γ ,σ),(γ ,σ))+D((γ ,σ),(γ ,τ))+D((γ ,τ),(γ ,id))+D((γ ,id),(γ ,id))
0 j j j j j j 0
≤
j+5 d(h,σ(h))+5 d(h,σ(h))+2(n j)+j
≤ −
h∈sXupp(cℓ) h∈sXupp(τ)
=2Cov(g,σ)+5 d(h,σ(h)).
h∈sXupp(σ)
⊓⊔
Proposition 1.4. Let (g,σ) G⋉Σ and let g′ G. Then,
∈ ∈
1
D((g,σ),(g′,id)) d(h,σ(h)).
≥ 2
h∈sXupp(σ)
Proof. TheproofisbyinductiononD =D((g,σ),(g′,id)). IfD =0thenσ =id,andweare
done. Assume that D >0. Let υ Υ be a generator such that D((g,σ)υ,(g′,id))=D 1.
∈ −
There exists u U such that either υ = (u,id) or υ = (e,< e,u >). If υ = (u,id) then by
∈
induction
1
D D((g,σ)υ,(g′,id)) d(h,σ(h)).
≥ ≥ 2
h∈sXupp(σ)
So assume that υ =(e,<e,u>). If σ(h) g,gu , then <g,gu>σ(h)=σ(h), and
6∈{ }
supp(σ) σ−1(g),σ−1(gu) =supp(<g,gu>σ) σ−1(g),σ−1(gu) .
\ \
(cid:8) (cid:9) (cid:8) (cid:9)
Since d(g,gu)=1,
d(h,σ(h))=d(g,σ−1(g))+d(gu,σ−1(gu))+ d(h,σ(h))
h∈sXupp(σ) h6∈{σ−1(Xg),σ−1(gu)}
d(g,gu)+d(gu,σ−1(g))+d(gu,g)+d(g,σ−1(gu))
≤
+ d(h,<g,gu>σ(h))
h6∈{σ−1(Xg),σ−1(gu)}
2+ d(h,<g,gu>σ(h)).
≤
h∈suppX(<g,gu>σ)
6
So by induction,
1
D =1+D((g,<g,gu>σ),(g′,id)) 1+ d(h,<g,gu>σ(h))
≥ 2
h∈suppX(<g,gu>σ)
1
d(h,σ(h)).
≥ 2
h∈sXupp(σ)
⊓⊔
2 The Mixer Chain on Z
We now consider the mixer chain on Z, with 1, 1 as the symmetric generating set. We
{ − }
denote by ω =(S ,σ ) the mixer chain on Z.
{ t t t }t≥0
Forω Z⋉ΣwedenotebyD(ω)thedistanceofωfrom(0,id)(withrespecttothegenerating
∈
set Υ, see Definition 1.2). Denote by D = D(ω ) the distance of the chain at time t from
t t
the origin.
Asstatedabove,weshowthatthe mixerchainonZhasrateofescape3/4. Infact,weprove
slightly stronger bounds on the distance to the origin at time t.
Theorem 2.1. Let D be the distance to the origin of the mixer chain on Z. Then, there
t
exist constants c,C >0 such that for all t 0, ct3/4 E[D ] Ct3/4.
t
≥ ≤ ≤
The proof of Theorem 2.1 is in Section 3.
For z Z, denote by X (z) = σ (z) z , the distance of the tile marked z to its origin at
t t
∈ | − |
time t. Define
X = X (z),
t t
z∈Z
X
whichis afinite sumfor anygivent. AsshowninPropositions1.3and1.4,X approximates
t
D up to certain factors.
t
For z Z define
∈ t
V (z)= 1 S =σ (z) .
t t t
{ }
j=0
X
V (z) is the number of times that the mixer visits the tile marked z, up to time t.
t
7
2.1 Distribution of Xt(z)
Proposition 2.2. For σ Σ define σ′ Σ by σ′(z)= σ( z) for all z Z. Then, for any
∈ ∈ − − ∈
t 1, ((S ,σ ),...,(S ,σ )) and (( S ,σ′),...,( S ,σ′)) have the same distribution.
≥ 1 1 t t − 1 1 − t t
Proof. Let ϕ be the permutation (not in Σ) defined by ϕ(x) = x for all x Z. Then,
− ∈
σ′ =ϕσϕ. Since ϕ2 =id, we get that (στ)′ =σ′τ′, σ′′ =σ and (σ−1)′ =(σ′)−1.
The proof is by induction on t. If t=1, then (S ,σ ) is uniformly distributed over the set
1 1
Υ= (1,id),( 1,id),(0,<0,1>),(0,<0, 1>) .
{ − − }
Since <0,1>′=<0, 1>, and id′ =id, the proposition is proved for t=1.
−
Let t > 1. Let (x,τ) Z⋉ Σ be any element, and let υ = (y,ρ) Υ be a generator.
∈ ∈
We have (x,τ)(y,ρ) = (x+y,xρ( x)τ). Since ρ id,<0,1>,<0, 1> , we have that
− ∈ { − }
(xρ( x)τ)′ =( x)ρ′xτ′. Since y =0 if and only if ρ=id, we get that ( y)ρy =ρ, so
− − 6 −
( x,τ′)( y,ρ′)=( (x+y),( x)ρ′xτ′)=( (x+y),(xρ( x)τ)′).
− − − − − −
Thus, we get that for any (z,σ),(x,τ) Z⋉Σ and any (y,ρ) Υ,
∈ ∈
(z,σ)=(x,τ)(y,ρ) if and only if ( z,σ′)=( x,τ′)( y,ρ′).
− − −
If (y,ρ) is uniformly distributed in Υ, then so is ( y,ρ′). Thus, since (S ,σ )−1(S ,σ )
t−1 t−1 t t
−
is uniformly distributed in Υ, ( S ,σ′ )−1( S ,σ′) is also uniformly distributed in Υ.
− t−1 t−1 − t t
By induction, ((S ,σ ),...,(S ,σ )) and (( S ,σ′),...,( S ,σ′ )), have the same
1 1 t−1 t−1 − 1 1 − t−1 t−1
distribution. The Markov property now implies the proposition.
⊓⊔
ByalazyrandomwalkonZ,werefertothe integervaluedprocessW ,suchthatW W
t t+1 t
−
arei.i.d. randomvariableswiththedistributionP[W W =1]=P[W W =1]=1/4
t+1 t t+1 t
− −
and P[W W =0]=1/2.
t+1 t
−
Lemma 2.3. Let t 0 and z Z. Let k 1 such that P[V (z)=k]>0. Then, conditioned
t
≥ ∈ ≥
on V (z) = k, the distribution of σ (z) z is the same as W +B, where W is a lazy
t t k−1 k
− { }
random walk on Z, and B is a random variable independent of W such that B 2.
k
{ } | |≤
Proof. Define inductively the following random times: T (z)=0, and for j 1,
0
≥
T (z)=inf t T (z)+1 : S =σ (z) .
j j−1 t t
{ ≥ }
8
Claim 2.4. Let T =T (0). For all ℓ such that P[T =ℓ]>0,
1
P σ (0)=1 T =ℓ =P σ (0)= 1 T =ℓ =1/4,
T T
−
(cid:2) (cid:12) (cid:3) (cid:2) (cid:12) (cid:3)
and (cid:12) (cid:12)
P σ (0)=0 T =ℓ =1/2,
T
(cid:2) (cid:12) (cid:3)
Proof. Note that S σ (0) = 1 and tha(cid:12)t for all 1 t < T, σ (0) = σ (0). Thus,
1 1 t 1
| − | ≤
σ (0)=σ (0) and S =S . So we have the equality of events
T−1 1 T−1 1
ℓ−1
T =ℓ = S =σ (0) S =S ,σ (0)=σ (0) S =σ (0) or σ (0)=S .
t t ℓ−1 1 ℓ−1 1 ℓ 1 ℓ 1
{ } { 6 } { } { }
t=1
\ \ \
Hence, if we denote = ℓ−1 S =σ (0) S =S ,σ (0)=σ (0) , then
E t=1{ t 6 t } { ℓ−1 1 ℓ−1 1 }
T T
P[T =ℓ]=P[ ] P S =σ (0) or σ (0)=S
ℓ 1 ℓ 1
E · E
1
=P[ ] .(cid:2) (cid:12)(cid:12) (cid:3) (2.1)
E · 2
Since the events S =0 and σ (0)=0 are disjoint and their union is the whole space,
1 1
{ } { }
we get that
P[σ (0)=0,T =ℓ]=P[ ,S =σ (0)=0]+P[ ,σ (0)=S =0]
T ℓ 1 ℓ 1
E E
=P[ ,σ (0)=0] P S =σ (0) S =S ,σ (0)=σ (0)=0
1 ℓ 1 ℓ−1 1 ℓ−1 1
E ·
+P[ ,S (0)=0] (cid:2)P σ (0)=S(cid:12) S =S =0,σ (0)=σ ((cid:3)0)
E 1 · ℓ (cid:12)1 ℓ−1 1 ℓ−1 1
1
=P[ ] . (cid:2) (cid:12)(cid:12) (2.(cid:3)2)
E · 4
Combining (2.1) and (2.2) we get that
1
P σ (0)=0 T =ℓ = .
T
2
(cid:2) (cid:12) (cid:3)
Finally, by Proposition 2.2, (cid:12)
P[σ (0)=1,T =ℓ]=P[ , S =σ (0)=1]
T ℓ ℓ
E
=P[σ (0)= 1,T =ℓ].
T
−
Since the possible values for σ (0) are 1,0,1, the claim follows.
T
− ⊓⊔
We continue with the proof of Lemma 2.3.
We have the equality of events V (z)=k = T (z) t<T (z) .
t k k+1
{ } { ≤ }
9
Let t ,t ,...,t ,t be such that
1 2 k k+1
P[T (z)=t ,...,T (z)=t ]>0,
1 1 k+1 k+1
and condition on the event = T (z)=t ,...,T (z)=t . Assume further that t
1 1 k+1 k+1 k
E { } ≤
t<t , so that V (z)=k. Write
k+1 t
k
σ (z) z =σ (z) σ (z)+ σ (z) σ (z)+σ (z) z. (2.3)
t − t − Tk(z) Tj(z) − Tj−1(z) T1(z) −
j=2
X
For1 j k 1denoteY =σ (z) σ (z). ByClaim2.4andtheMarkovproperty,
≤ ≤ − j Tj+1(z) − Tj(z)
conditioned on , Y are independent with the distribution P[Y =1 ]=P[Y = 1 ]=
j j j
E { } |E − |E
1/4andP[Y =0 ]=1/2. Soconditionedon , k−1Y hasthesamedistributionofW .
j |E E j=1 j k−1
Finally, σ (z) σ (z) 1, and σ (z) Pz 1. Since conditioned on , σ (z)
| t − Tk(z) | ≤ | T1(z) − | ≤ E t −
σ (z), and σ (z) z are independent of Y , this completes the proof of the lemma.
Tk(z) T1(z) − { j}
⊓⊔
Corollary 2.5. There exist constants c,C >0 such that for all t 0 and all z Z,
≥ ∈
cE[ V (z)] 2P[V (z) 1] E[X (z)] CE[ V (z)]+2P[V (z) 1].
t t t t t
− ≥ ≤ ≤ ≥
p p
Proof. Let W be alazy randomwalk onZ. Note that 2W hasthe samedistributionas
t t
{ } { }
S′ where S′ is a simple random walk on Z. It is well known (see e.g. [3]), that there
{ 2t} { t}
exist universal constants c ,C >0 such that for all t 0,
1 1
≥
c √t E[S′ ]=2E[W ] C √t.
1 ≤ | 2t| | t| ≤ 1
By Lemma 2.3, we know that for any k 0,
≥
E[W ] 2 E X (z) V (z)=k+1 E[W ]+2.
k t t k
| | − ≤ ≤ | |
(cid:2) (cid:12) (cid:3)
Thus, summing over all k, there exists con(cid:12)stants c ,C >0 such that
2 2
c E[ V (z)] 2P[V (z) 1] E[X (z)] C E[ V (z)]+2P[V (z) 1].
2 t t t 2 t t
− ≥ ≤ ≤ ≥
p p
⊓⊔
Lemma 2.6. Let S′ be a simple random walk on Z started at S′ =0, and let
{ t} 0
t
L (z)= 1 S′ =z .
t j
j=0
X (cid:8) (cid:9)
10